Maths 4 Marks Questions

Graph of the equation 2x + 3y = 12:

We have,

2x + 3y = 12

⇒ 2x = 12 - 3y

$\Rightarrow\text{x}=\frac{12-3\text{y}}{2}\ ...\text{(i)}$

Putting y = 4, we get $\text{x}=\frac{12-3\times4}{2}=0$

Putting y = 2, we get $\text{x}=\frac{12-3\times2}{2}=3$

Thus, we have the following table for the points on the line 2x + 3y = 12:

Ploting points A(0, 4), B(3, 2) on the graph paper and drawing a line passing through them, we obtain graph of the equation 2x + 3y = 12.

Graph of the equation x - y = 1:

We have,

x - y = 1

⇒ x = 1 + y

Putting y = 0, we get x = 1 + 0 = 1

Putting y = -1, we get x = 1 - 1 = 0

Thus, we have the following table for the points on the line x - y = 1:

Ploting points C(1, 0) and D(0, -1) on the same graph paper and drawing a line passing through them, we obtain the graph of the line represented by the equation x - y = 1.

Clearly, two lines intersect at A(3, 2).

The graph of line 2x + 3y = 12 intersect with y-axis at B(0, 4) and the graph of the line x - y = 1 intersect with y-axis at C(0, -1).

So, the vertices of the triangle formed by the two straight lines and y-axis are A(3, 2), B(0, 4) and C(0, -1).

Now,

$\text{Area of }\triangle\text{ABC}=\frac{1}{2}(\text{Base}\times\text{Height})$

$=\frac{1}{2}(\text{BC}\times\text{AM})$

$=\frac{1}{2}(5\times3)$

$=\frac{15}{2}\text{sq.units.}$

⇒ x - 2 = 3(y - 3)

⇒ x - 2 = 3y - 9

⇒ x = 3y - 9 + 2

⇒ x = 3y - 7 ...(i)